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$\int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{3}\left(e^{75}-e\right)$
Let $I=\int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x$
Let $\frac{3}{x}=t$
$\Rightarrow-\frac{3}{x^2} d x=d t \Rightarrow \frac{1}{x^2} d x=-\frac{1}{3} d t$
When $x=3 \Rightarrow t=1$
$\begin{aligned} & x=\frac{1}{25} \Rightarrow t=75 \\ & \therefore I=\int_{75}^1 e^t\left(-\frac{1}{3} d t\right)=-\frac{1}{3}\left[e^t\right]_{75}^1 \\ & =-\frac{1}{3}\left[e-e^{75}\right]\end{aligned}$
$\Rightarrow \quad I=\frac{1}{3}\left[e^{75}-e\right]$
Let $\frac{3}{x}=t$
$\Rightarrow-\frac{3}{x^2} d x=d t \Rightarrow \frac{1}{x^2} d x=-\frac{1}{3} d t$
When $x=3 \Rightarrow t=1$
$\begin{aligned} & x=\frac{1}{25} \Rightarrow t=75 \\ & \therefore I=\int_{75}^1 e^t\left(-\frac{1}{3} d t\right)=-\frac{1}{3}\left[e^t\right]_{75}^1 \\ & =-\frac{1}{3}\left[e-e^{75}\right]\end{aligned}$
$\Rightarrow \quad I=\frac{1}{3}\left[e^{75}-e\right]$
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